Is any closed-form representation known for the sum $\sum\limits_{n=1}^{\infty}\frac{\mu(n)\log n}{n^2}$, where $\mu(n)$ is the Möbius $\mu$-function?


You can use a series for the reciprocal Riemann $\zeta$-function: $\zeta(s)^{-1}=\sum\limits_{n=1}^{\infty}\mu(n)n^{-s}$, then take a derivative with respect to $s$ and let $s=2$. This gives you $\sum\limits_{n=1}^{\infty}\frac{\mu(n)\log n}{n^2}=\frac{\zeta'(2)}{\zeta(2)^2}$. It is well-known that $\zeta(2)=\frac{\pi^2}{6}$, and a closed form of the derivative is given in OEIS A073002:

$\zeta'(2)=\frac{\pi^2}{6}\left(\gamma+\log(2\pi)-12 \log A\right)$,

where $\gamma$ is the Euler-Mascheroni constant, and $A$ is the Glaisher-Kinkelin constant. Taking it all together, the result is:

$\sum\limits_{n=1}^{\infty}\frac{\mu(n)\log n}{n^2}=\frac{6}{\pi^2}\left(\gamma+\log(2\pi)-12 \log A\right)$.

  • 13
    $\begingroup$ Very nice answer! Mathematica was not able to calculate this. $\endgroup$ – Zakharia Stanley May 5 '13 at 4:10
  • $\begingroup$ @ZakhariaStanley That says something about humans, no? $\endgroup$ – user285523 Nov 8 '15 at 3:48

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